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Chapter 8: Problem 58

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

Short Answer

Expert verified
The dimensions of the rectangle could either be 8 feet by 12 feet or 12 feet by 8 feet.

Step by step solution

01

Equation Formulation

First, formulate the two equations from the given condition: The perimeter: \(2l + 2w = 40\) and the area: \(l \times w = 96\).
02

Equation Simplification

Simplify the perimeter equation \(2l + 2w = 40\) to get the equation in terms of one variable. For instance, rearranging for w, we get \(w = 20 - l)\).
03

Substitution Method

Next, use the equation \(w = 20 - l\) derived from the perimeter equation and substitute \(w\) in the area equation \(l \times w = 96\) to get an equation in term of one variable. This results in the equation \(l \times (20 - l) = 96\).
04

Solve for one variable

Solve the quadratic equation \(l \times (20 - l) = 96\) which simplifies to \(l^2 -20l + 96 = 0\). Factoring the quadratic gives us \((l-12)(l-8) = 0\). Solving for \(l\) won't yield negative values because it's a length, so \(l\) can either be 8 or 12.
05

Solve for the second variable

Substitute each of the obtained \(l\) values in equation \(w = 20 - l\) to obtain the corresponding \(w\) values. When \(l=8\), \(w=12\), and when \(l=12\), \(w=8\). So the length and width could be (8,12) or (12,8).

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